A Borsuk Theorem for Antipodal Links and a Spectral Characterization of Linklessly Embeddable Graphs1 LÁszlÓ LovÁsz2 and Alexander Schrijver3

A Borsuk theorem for antipodal links and a spectral characterization of linklessly embeddable graphs1 László Lovász2 and Alexander Schrijver3

Abstract. For any undirected graph G, let µ(G) be the graph parameter introduced by Colin de Verdi`ere. In this paper we show that µ(G) ≤ 4 if and only if G is linklessly embeddable (in 3 R ). This forms a spectral characterization of linklessly embeddable graphs, and was conjectured by Robertson, Seymour, and Thomas. A key ingredient is a Borsuk-type theorem on the existence of a pair of antipodal linked (k − 1)- 2k−1 spheres in certain mappings φ : S2k → R . This result might be of interest in its own right. We also derive that λ(G) ≤ 4 for each linklessly embeddable graph G = (V, E), where λ(G) is the graph paramer introduced by van der Holst, Laurent, and Schrijver. (It is the largest dimension V of any subspace L of R such that for each nonzero x ∈ L, the positive support of x induces a nonempty connected subgraph of G.)

1. Introduction

Motivated by estimating the maximum multiplicity of the second eigenvalue of Schrödin- ger operators, Colin de Verdière [4] (cf. [5]) introduced an interesting new invariant µ(G) for graphs G, based on spectral properties of matrices associated with G. He showed that the invariant is monotone under taking minors (that is, if H is a minor of G then µ(H) ≤ µ(G)), that µ(G) ≤ 1 if and only if G is a disjoint union of paths, that µ(G) ≤ 2 if and only if G is outerplanar, and that µ(G) ≤ 3 if and only if G is planar. In this paper we show that µ(G) ≤ 4 if and only if G is linklessly embeddable in R3. An embedding of a graph G in R3 is called linkless if each pair of disjoint circuits in G are unlinked closed curves in R3 (for our ourposes, the following definition of linking suffices: two disjoint curves are unlinked, if there is a mapping of the unit disc into R3 such that its boundary is mapped onto the first curve and the image of the disc is disjoint from the second). G is linklessly embeddable if G has a linkless embedding in R3. (‘Embedding’ presumes that vertices and edges have disjoint images.) Our result was conjectured by Robertson, Seymour, and Thomas [9], and has the following context. Robertson, Seymour, and Thomas [10] showed that a graph G is linklessly embeddable if and only if G does not have any of the seven graphs in the Petersen family as a minor — the Petersen family consists of all graphs obtainable from K6 by so-called ∆Y- and Y∆-operations (it includes the Petersen graph). (Y∆ means deleting a vertex of degree 3, and making its three neighbours mutually adjacent. ∆Y is the reverse operation (applied to a triangle).)

11991 Mathematics Subject Classiﬁcation: primary: 05C10, 05C50, 57M15, secondary: 05C50, 57N15 2Department of Computer Science, Yale University, New Haven, Connecticut 06520, U.S.A. email: [email protected] 3CWI, Kruislaan 413, 1098 SJ Amsterdam, The Netherlands and Department of Mathematics, University of Amsterdam, Plantage Muidergracht 24, 1018 TV Amsterdam, The Netherlands. Research partially done while visiting the Department of Computer Science at Yale University. email: [email protected]

1 Since any of the graphs G in the Petersen family has µ(G) ≥ 5 (Bacher and Colin de Verdière [1]), it follows that if µ(G) ≤ 4 then G is linklessly embeddable. So we prove the reverse implication. Thus, next to the combinatorial characterization (in terms of minors) of the (topologically defined) class of linklessly embeddable graphs, there is a spectral characterization. Our proof method also applies to a related parameter called λ(G), introduced by van der Holst, Laurent, and Schrijver [7]. It is defined as follows. Let G = (V, E) be a graph. Call V a linear subspace L of R a representation of G if for each x ∈ L, supp+(x) is nonempty and G|supp+(x) is a connected graph. Here we use the following notation. If U ⊆ V then G|U is the subgraph of G induced by U (and G − U = G|(V \ U)). For any vector x ∈ RV , supp(x) is the support of x, that is, supp(x) := {v ∈ V |x(v) 6= 0}. The positive support is supp+(x) := {v ∈ V |x(v) > 0} and the negative support is supp−(x) := {v ∈ V |x(v) < 0}. Now by definition, λ(G) is the maximum dimension of any representation L of G. It is easy to see that λ(G) is monotone under taking minors. Moreover, if G is a clique sum of graphs G1 and G2 (that is, if G arises from G1 and G2 by identifying a clique in G1 and G2), then λ(G) = max{λ(G1), λ(G2)}. In [7] it is shown that λ(G) ≤ 1 if and only if G is a forest, that λ(G) ≤ 2 if and only if G is series-parallel, and that λ(G) ≤ 3 if and only if G arises by taking clique sums and subgraphs from planar graphs. In this paper we show that λ(G) ≤ 4 if G is linklessly embeddable — hence, more generally, if G arises by taking clique sums and subgraphs from linklessly embeddable graphs. We do not know if the reverse implication holds. A key ingredient in our proof is a Borsuk-type theorem on the existence of antipodal links, which we formulate and prove in Section 2. We derive it from an extension of a theorem of Bajmóczy and Bárány [2] establishing a polyhedral form of Borsuk’s antipodal theorem. In Section 3 we derive that λ(G) ≤ 4 for linklessly embeddable graphs G. In Section 4 we give some preliminaries on the Colin de Verdère parameter µ(G), and after that, in Section 5, we show that µ(G) ≤ 4 for linklessly embeddable graphs G. Finally, in Section 6 we consider a number of open questions related to µ(G) and λ(G).

2. A Borsuk-type theorem for antipodal links

Let P be a convex polytope in Rn. We say that two faces F and F 0 are antipodal if there exists a nonzero vector c in Rn such that the linear function cT x is maximized by every point of F and minimized by every point of F 0 (Figure 1). So F and F 0 are antipodal if and only if F − F 0 is contained in a face of P − P . Call a continuous map φ of a cell complex into Rm generic if the images of a k-face and an l-face intersect only if k + l ≥ m, and for k + l = m they have a finite number of intersection points, and at these points they intersect transversally. (In this paper, faces are relatively open.) For any convex polytope P in Rn, let ∂P denote its boundary. The following theorem extends a result of Bajmóczy and Bárány [2]. (The difference is that their theorem concludes that φ(F ) ∩ φ(F 0) is nonempty. Their proof uses Borsuk’s theorem. We give an independent proof.)

2 Theorem 1. Let P be a full-dimensional convex polytope in Rn and let φ be a generic continuous map from ∂P to Rn−1. Then there exists a pair of antipodal faces F and F 0 with dim(F ) + dim(F 0) = n − 1 such that |φ(F ) ∩ φ(F 0)| is odd.

Proof. We prove a more general fact. Call two faces parallel if their projective hulls have a nonempty intersection that is contained in the hyperplane at inﬁnity. So faces F and F 0 are parallel if and only if their aﬃne hulls are disjoint while F − F and F 0 − F 0 have a nonzero vector in common. (Note that two antipodal faces are parallel if dim(F ) + dim(F 0) ≥ n.)

Figure 1: Parallel and non-parallel antipodal faces

Now it suﬃces to show

(1) Let P be a full-dimensional convex polytope in Rn having no parallel faces and let φ be a generic continuous map from ∂P to Rn−1. Then

0 X |φ(F ) ∩ φ(F )|

is odd, where the summation extends over all antipodal pairs {F, F 0} of faces. (It would be enough to sum over all antipodal pairs {F, F 0} with dim(F )+dim(F 0) = n−1.) To see that it suffices to prove (1), it is enough to apply a random projective transformation close to the identity. To be more precise, assume that we have a polytope P that has parallel faces. For every pair (E, E0) of faces whose affine hulls intersect, choose a (fi- 0 nite) point pEE0 in the intersection of the affine hulls. For every pair (E, E ) of faces whose projective hulls intersect, choose an infinite point qEE0 in the intersection of their projective hulls. Let H be a finite hyperplane having all the points pEE0 on one side, and avoiding all the points qF F 0 . Apply a projective transformation that maps H onto the hyperplane at infinity, to get a new polytope P 0. It is clear that P 0 has no parallel faces, and it is easy to argue that every pair of faces that are antipodal in P 0 correspond to antipodal faces in P . Hence (1) implies the theorem. We now prove (1). Let P be a convex polytope in Rn having no parallel faces. For any

3 two faces F, F 0, denote F ≤ F 0 if F ⊆ F 0. Then: (2) (i) if A and B are antipodal faces, then A − B is a face of P − P , with dim(A − B) = dim(A) + dim(B); (ii) if F is a face of P − P , then there exists a unique pair A, B of antipodal faces with A − B = F ; (iii) for any two pairs A, B and A0, B0 of antipodal faces one has: A−B ≤ A0−B0 if and only if A ≤ A0 and B ≤ B0. This gives the following observation: (3) For every pair of faces A and B with dim(A) + dim(B) = n − 2, the number of antipodal pairs {F, F 0} of faces with A ≤ F and B ≤ F 0 and dim(F )+dim(F 0) = n − 1 is 0 or 2. To see (3), it is clear that if A and B are not antipodal, then this number is 0. Suppose that they are antipodal. Then the number is 2, since by (2), it is equal to the number of facets of P − P incident with the (n − 2)-face A − B. To prove (1), we use a “deformation” argument. The statement is true for the following mapping φ: pick a point q very near the center of gravity of some facet F (outside P ), and project ∂P from q onto the hyperplane H of F . Then the only nontrivial intersection is that the image of the (unique) vertex of P farthest from H is contained in F . Now we deform this map to φ. We may assume that the images of two faces E and E0 with dim(E)+dim(E0) ≤ n−3 never meet; but we have to watch when φ(A) passes through φ(B), where A and B are faces with dim(A) + dim(B) = n − 2. But then |φ(F ) ∩ φ(F 0)| changes exactly when A ⊆ F and B ⊆ F 0. By (3), this does not change the parity. This proves (1), and hence the theorem.

For any polytope P , let (P )k denote its k-skeleton. Two disjoint images A and B of (d − 1)-spheres in R2d−1 are said to have an odd linking number if the image of A can be extended to the image of a d-ball with an odd number of transveral intersections with the image of B (and no other intersections). So having an odd linking number implies being linked.

Corollary 1.1. Let P be a full-dimensional convex polytope in R2k+1 and let φ be an 2k−1 0 embedding of (P )k−1 into R . Then there exists a pair of antipodal k-faces F and F such that φ(∂F ) and φ(∂F 0) have an odd linking number.

Proof. First we extend φ with a last coordinate equal to 0, to obtain an embedding ψ of 2k 2k (P )k−1 into R . Next we extend ψ to a generic mapping ∂P → R , in such a way that ψ(x) has last coordinate positive if x ∈ ∂P \ (P )k−1. By Theorem 1, P has two antipodal faces F and F 0 such that dim(F ) + dim(F 0) = 2k and |ψ(F ) ∩ ψ(F 0)| is odd. If dim(F ) ≤ k − 1, then the last coordinate of each point in ψ(F ) is 0, while the last coordinate of each point in ψ(F 0) is positive (as dim(F 0) ≥ k). So dim(F ) ≥ k and similarly dim(F 0) ≥ k. Therefore, dim(F ) = dim(F 0) = k. 0 Then the boundaries of F and F are (k − 1)-spheres S1 and S2, mapped disjointly into R2k−1, and the mappings extend to mappings of the k-balls into the “upper” halfspace of

4 R2k, so that the images of the balls intersect at an odd number of points. But this implies that the images of the spheres are linked. In fact, if they were not linked, then there exists an extension of the map of ∂F to a continuous mapping ψ0 of F into R2k such that the image of every point in the interior of F has last coordinate equal to 0, and ψ0(F ) intersects ψ(∂F 0) transversally in an even number of points. We can extend the map of ∂F 0 to a continuous mapping ψ0 of F 0 into R2k such that the image of every point in the interior of F has a negative last coordinate 0. Then we get two maps of the k-sphere into R2k with an odd number of transversal intersection points, which is impossible. This contradiction completes the proof.

3. λ(G) ≤ 4 for linklessly embeddable graphs G

In this section we focus on the graph parameter λ(G) introduced in [7], and show that λ(G) ≤ 4 for linklessly embeddable graphs. The proof method also serves as an introduction to the methods used in proving that µ(G) ≤ 4 for linklessly embeddable graphs. We gave the deﬁnition of λ(G) in Section 1.

Theorem 2. If G is linklessly embeddable, then λ(G) ≤ 4.

Proof. Let G be linklessly embedded in R3, and suppose that λ(G) ≥ 5. Then there is a V 5-dimensional subspace L of R such that G|supp+(x) is nonempty and connected for each nonzero x ∈ L. 0 0 Call two elements x and x of L equivalent if supp+(x) = supp+(x ) and supp−(x) = 0 supp−(x ). The equivalence classes decompose L into a centrally symmetric complex P of pointed polyhedral cones. Choose a suﬃciently dense set of vectors of unit length from every cone in P, in a centrally symmetric fashion, and let P be the convex hull of these vectors. Then P is a 5-dimensional centrally symmetric convex polytope such that every face of P is contained in a cone of P. 3 0 We deﬁne an embedding φ of (P )1 in R . For each vertex v of P , we choose a node v 3 0 in supp+(v), end we let φ(x) be a point in R very near v . For each edge e = uv of P , we 0 0 0 choose a path e connecting u and v in G|supp+(x), where x is an interior point of e. (By our construction, supp+(x) is independent of the choice of x, and contains both supp+(u)

P +

v x 0 e

u +

- -

Figure 2: Constructing an embedding of (P )1

5 0 and supp+(v).) Then we map e onto a Jordan curve connecting φ(u) and φ(v) very near e . Clearly we can choose the images of the vertices and edges so that this map φ is one-to-one (Fig. 2). Then by Corollary 1.1, P has two antipodal 2-faces F and F 0 such that the images of their boundaries are linked. Since P is centrally symmetric, there is a facet D of P such that F ⊆ D and F 0 ⊆ −D. Let y be a vector in the interior of D. Then the images of 0 ∂F and ∂F are very near subgraphs spanned by supp+(y) and supp−(y), respectively, and hence some cycle of G spanned by supp+(y) must be linked with some cycle in supp−(y), a contradiction.

Corollary 2.1. If G is obtained from linklessly embedded graphs by taking clique sums and subgraphs, then λ(G) ≤ 4.

Proof. Directly from Theorem 2 and the fact that the class of graphs G with λ(G) ≤ 4 is closed under taking clique sums and subgraphs ([7]).

4. The Colin de Verdi`ere parameter µ(G)

We now go over to the Colin de Verdi`ere parameter µ(G), to which we ﬁrst give some background. Let G = (V, E) be an undirected graph, which we assume without loss of generality to have vertex set {1, . . . , n}. Then µ(G) is the largest corank of any symmetric real-valued n × n matrix M = (mi,j) satisfying: (4) (i) M has exactly one negative eigenvalue, of multiplicity 1,

(ii) for all i, j with i 6= j: mi,j < 0 if i and j are adjacent, and mi,j = 0 if i and j are nonadjacent,

(iii) there is no nonzero symmetric n × n matrix X = (xi,j) such that MX = 0 and such that xi,j = 0 whenever i = j or mi,j 6= 0.

There is no condition on the diagonal entries mi,i. (The corank corank(M) of a matrix M is the dimension of its kernel.) Condition (4)(iii) is called the Strong Arnold Property (or Strong Arnold Hypothesis). There exist matrices M satisfying (4), for which ker(M) is a not a representation of G; that is, for which there exist x ∈ ker(M) with G|supp+(x) disconnected. (Otherwise µ(G) ≤ λ(G) would follow, which is not true.) The Petersen graph provides an example. Let A be the adjacency matrix of the Petersen graph and let M = I − A. Let e and e0 be V two edges at distance 2, and deﬁne a vector qee0 ∈ R as 1 on the endnodes of e, −1 on 0 the endnodes of e , and 0 elsewhere. Then qee0 ∈ ker(M), and it is easy to see that in fact ker(M) is generated by these vectors. Now if e, e0 and e00 are three edges that are mutually 0 00 at distance 2, then qee + qee is a vector in ker(M) with supp−(q) having two components (Fig. 3). We shall see that this is as bad as it ever gets. The following lemma extends a lemma in [6]; the methods we use to prove it are close to those used in [8]. For any graph G = (V, E) and U ⊆ V , let N(U) be the set of vertices in V \ U that are adjacent to at least one vertex in U. For any V × V matrix and I, J ⊆ V , let MI×J denote

6 -1

-1

-1 -1 -1 -1

1 1 2 2

Figure 3: A vector on the Petersen graph with disconnected positive support

the submatrix induced by the rows in I and columns in J, and let MI := MI×I . For any I vector z ∈ R and J ⊆ I, let zJ be the subvector of z induced by the indices in J.

Lemma 1. Let G be a connected graph, let M be a matrix satisfying (4), and let x be a vector in ker(M) with G|supp+(x) disconnected. Then there are no edges connecting supp+(x) and supp−(x), and each component K of G|supp(x) satisﬁes N(K) = N(supp(x)). Proof. Let z be a positive eigenvector of M (by the Perron-Frobenius theorem, z is unique up to scaling, and belongs to the smallest eigenvalue of M). Let I and J be two components of G|supp+(x). Let L := supp−(x). As Mx = 0, we have

(5) MI×I xI + MI×LxL = 0, MJ×J xJ + MJ×LxL = 0.

T T V Let λ := zI xI /zJ xJ . Deﬁne y ∈ R by: yi := xi if i ∈ I, yi := −λxi if i ∈ J, and yi := 0 if T T T i 6∈ I ∪ J. Then z y = zI xI − λzJ xJ = 0. Moreover, one has (since MI×J = 0): T T T T 2 T (6) y My = yI MI yI + yJ MJ yJ = xI MI xI + λ xJ MJ xJ = T 2 T −xI MI×LxL − λ xJ MJ×LxL ≤ 0

(using (5)), since MI×L and MJ×L are nonpositive, and since xI > 0, xJ > 0 and xL < 0. Now zT y = 0 and yT My ≤ 0 imply that My = 0 (as M is symmetric and has exactly one negative eigenvalue, with eigenvector z). Therefore, y ∈ ker(M). By (4)(i), for distinct u, v, entry Mu,v of M is negative if and only if u and v are adjacent. This implies that each vertex in V \ supp(y) adjacent to supp+(y) = I is also adjacent to supp−(y) = J, and conversely; that is, N(I) = N(J). Also x − y belongs to ker(M). Hence, again, any vertex v ∈ V \ supp(x − y) is adjacent to supp−(x−y) if and only if v is adjacent to supp+(x−y). As supp+(x−y) = supp+(x)\I and as I is a component of G|supp+(x), no vertex in I is adjacent to supp+(x − y). Hence no vertex in I is adjacent to supp−(x − y) = supp−(x). As I is an arbitrary component of G|supp+(x), it follows that there is no edge connecting supp+(x) and supp−(x). So each component of G|supp(x) is a component of either G|supp+(x) or G|supp−(x).

7 Since N(I) = N(J) for any two components I, J of G|supp+(x), and similarly, N(I) = N(J) for any two components I, J of G|supp−(x), and since N(supp+(x)) = N(supp−(x)), we have that N(I) = N(supp(x)) for any component I of G|supp(x).

5. µ(G) ≤ 4 for linklessly embeddable graphs

Theorem 3. A graph G is linklessly embeddable if and only if µ(G) ≤ 4.

Proof. By the results of [10] and [1] it suffices to show that µ(G) ≤ 4 if G is linklessly embeddable. Let G be linklessly embeddable and suppose that µ(G) ≥ 5. By the results of [10] we may assume that G is ‘flatly embedded’ in R3; that is, for each circuit C in G there exists an open disk (“panel”) D in R3 with the property that D is disjoint from G and has boundary C. (Any flat embedding is also linkless, but the reverse is generally not true. But in [10] it has been shown that if G has a linkless embedding it also has a flat embedding.) We take a counterexample G with a minimum number of vertices. Then G is 4- connected. For suppose that G has a minimum-size vertex cut U with |U| ≤ 3. Consider any component K of G − U. Then the graph G0 obtained from G − K by adding a clique on U is a linkless embeddable graph again, because, if |U| ≤ 2, G0 is a minor of G, and if |U| = 3, G0 can be obtained from a minor of G by a Y∆-operation. As G0 has fewer vertices than G, we have µ(G0) ≤ 4. As this is true for each component K, G is a a subgraph of a clique sum of graphs G0 with µ(G0) ≤ 4, with cliques of size at most 3, and hence by the results of [8], µ(G) ≤ 4. Let M be a matrix satisfying (4) with corank(M) = 5. We proceed as in the proof 0 0 of Theorem 2. Call two elements x and x of ker(M) equivalent if supp+(x) = supp+(x ) 0 and supp−(x) = supp−(x ). The equivalence classes decompose ker(M) into a centrally symmetric complex P of pointed polyhedral cones. Call a cone f of P broken if G|supp+(x) is disconnected for any x ∈ f. To study broken cones, we first remark:

(7) for each x ∈ ker(M) with G|supp+(x) disconnected, G|supp(x) has exactly three components, say K1, K2, and K3, with K1 ∪K2 = supp+(x) and K3 = supp−(x), and with N(Ki) = V \ supp(x) for i = 1, 2, 3. This follows directly from Lemma 1, using the 4-connectivity of G and the fact that G has no K4,4-minor (as K4,4 is not linklessly embeddable (cf. [10])). Now (7) gives:

(8) any broken cone f is 2-dimensional.

Indeed, choose x ∈ f, and let K1, K2, and K3 be as in (7). Consider any y ∈ f. As supp(y) = supp(x) we have that MKi yKi = 0 for i = 1, 2, 3. As MKi xKi = 0 and as xKi is fully positive or fully negative, we know by the Perron-Frobenius theorem that yKi = λixKi for some λi > 0 (i = 1, 2, 3). Moreover, for the positive eigenvector z of M we have that zT y = zT x = 0. Conversely, any vector y ∈ RV with zT y = 0 and supp(y) = supp(x) and for which there exist λ1, λ2, λ3 > 0 with yKi = λixKi for i = 1, 2, 3, belongs to f, since

8 it belongs to ker(M). This follows from the fact that zT y = 0 and yT My = 0. So f is 2-dimensional, proving (8). Now choose a suﬃciently dense set of vectors of unit length from every cone in P, in a centrally symmetric fashion, and let P be the convex hull of these vectors. Then P is a 5-dimensional centrally symmetric convex polytope such that every face of P is contained in a cone of P. We choose the vectors densely enough such that every face of P contains at most one edge that is part of a 2-dimensional cone in P. We call an edge of P broken if it is contained in a broken cone in P. 3 We deﬁne an embedding φ of the 1-skeleton (P )1 of P in R . We map each vertex x of P to a point φ(x) near supp+(x), and we map any unbroken edge e = xy of P to a curve connecting φ(x) and φ(y) near G|supp+(z), where z ∈ e. We do this in such a way that the mapping is one-to-one. Consider next a broken edge e of P . Choose x ∈ e, let K1, K2,and K3 be as in (7), and let T := N(supp(x)). Then

3 (9) there is a curve C in R \ G connecting K1 and K2 such that there is no pair of disjoint linked circuits A in G|(K1 ∪ K2 ∪ T ) ∪ C and B in G|(K3 ∪ T ).

To see this, let H be the flatly embedded graph obtained from G by contracting Ki to one vertex vi (i = 1, 2, 3). It suffices to show that there is a curve C connecting v1 and v2 such that the graph H ∪ C is linklessly embedded. (Indeed, having C with H ∪ C linklessly embedded, we can decontract each Ki slightly, and make C connecting two arbitrary points in K1 and K2. Consider a circuit A in G|(K1 ∪ K2 ∪ T ) ∪ C and a circuit B in G|(K3 ∪ T ) 0 0 disjoint from A. Contracting K1, K2, and K3, we obtain disjoint cycles A and B in H ∪C. (A cycle is an edge-disjoint union of circuits.) As H ∪ C is linklessly embedded, A0 and B0 are unlinked. Hence A and B are unlinked.) Now H|T is a Hamiltonian circuit on T , or part of it. Otherwise, H|T would contain, as a minor, a graph on four vertices that is either a K1,3 or a triangle with an isolated vertex. In both cases, it implies that H has a minor in the Petersen family, which is not possible since H is linklessly embedded. So H is isomorphic to the complete bipartite graph K3,|T |, with some edges on T added forming part (or whole) of a Hamiltonian circuit on T . As H is flatly embedded, for each edge t1t2 of H|T there is an open disk (“panel”) with boundary the triangle t1t2v3, in such a way that the panels are pairwise disjoint (by Böhme’s lemma [3] (cf. [11], [10])). Since the union of H|(K3 ∪ T ) with the panels is contractible, there is a curve C from v1 to v2 not intersecting any panel. This curve has the required properties, showing (9). We now define φ on e close to a curve in G|(K1 ∪ K2) ∪ C, again so that it is one-to-one. We do this for each broken edge e, after which the construction of φ is finished. Then by Corollary 1.1, there are two antipodal 2-faces F and F 0 such that the images of their boundaries are linked. Since P is centrally symmetric, there is a facet D of P such that F ⊆ D and F 0 ⊆ −D. Let y be a vector in the interior of D. Then ∂F and ∂F 0 have 0 image in supp+(y) and supp−(y) respectively. If ∂F and ∂F do not contain any broken edge, then it would follow that G has two disjoint linked circuits — a contradiction. So we can assume that ∂F contains a broken edge e. Then it is the only broken edge in ∂F , since by our construction, ∂D contains at most one edge of P that is part of a 2-dimensional cone f in P. So f is broken. Moreover, ∂F 0 does not contain any broken

9 edge. For suppose that ∂F 0 contains broken edge e0 of P . Then e0 is part of a broken 2-dimensional cone f 0 in P, and hence f 0 = −f (since D is incident with at most one edge that is part of a 2-dimensional cone of P). However, as f is broken, −f is not broken, since G|supp−(x) is connected for any x ∈ f (by (7)). Choose x ∈ f. and consider the partition of V into K1, K2, K3, and T as above, with supp+(x) = K1 ∪ K2 and supp−(x) = K3. Then K1 ∪ K2 ⊆ supp+(y) and K3 ⊆ supp−(y), and hence supp+(y) ⊆ K1 ∪ K2 ∪ T and supp−(y) ⊆ K3 ∪ T . So the image of ∂F is close to G|(K1 ∪ K2 ∪ T ) ∪ C, where C is the curve constructed for the broken edge e of P , and 0 the image of ∂F is close to G|(K3 ∪ T ). This contradicts (9).

Note that in this proof, the Strong Arnold Property is hardly used. Also the Lemma remains true without the Strong Arnold Property. In fact, the above shows that for a 4-connected linklessly embeddable graph G, each matrix M satisfying (4)(i) and (ii) has corank(M) ≤ 4.

6. Some open questions

A ﬁrst question that comes up is if one can prove that a graph G is linklessly embeddable if µ(G) ≤ 4 directly, that is, without using the Robertson-Seymour-Thomas theorem. In other words, does the nonexistence of a matrix M of corank 5 satisfying (4) imply, in a direct way, the existence of a linkless embedding? Secondly, the above does not complete the characterization of graphs G satisfying λ(G) ≤ 4. Each graph G obtainable from linklessly embeddable graphs by taking clique sums and subgraphs, satisﬁes λ(G) ≤ 4, but it is an open question if the reverse also holds. To answer this question, one might investigate the class of minor-minimal graphs that can- not be obtained from linklessly embeddable graphs by taking clique sums and subgraphs. It is not known what the full list of these graphs is. In [7] the following graphs are shown to be minor-minimal with respect to the property λ(G) ≥ 5. First G = K6 (all other graphs G in the Petersen family satisfy λ(G) ≤ 4). Next consider the graph V8 with vertices v1, . . . , v8, with vi and vj adjacent if and only if |i−j| ∈ {1, 4, 7}. (It was shown by Wagner [12] that a graph G can be obtained from planar graphs by taking clique sums and subgraphs if and only if G has no K5- or V8-minor. So K5 and V8 are the only minor-minimal graphs G with λ(G) ≥ 4.) 0 Let V9 arise from V8 by adding an extra vertex v0, adjacent to v2, v4, v6, v7, v8. Similarly, 00 let V9 arise from V8 by adding an extra vertex v0 adjacent to v2, v3, v5, v7, v8. It is shown 0 00 in [7] that V9 and V9 are minor-minimal graphs G with λ(G) ≥ 5. 0 00 The graphs V9 and V9 are also minor-minimal graphs not obtainable from linklessly embeddable graphs by taking clique sums and subgraphs. This can be seen as follows. 0 00 Since λ(V9) = λ(V9 ) = 5, it follows from Corollary 2.1 that these two graphs indeed are not obtainable in such a way. Moreover, to see that they are minor-minimal, observe that 0 00 deleting or contracting any edge of V9 or V9 produces a graph that has a vertex whose deletion makes the graph a clique sum of planar graphs. Since the class of graphs G with λ(G) ≤ 4 is closed under taking ∆Y operations (not 0 under Y∆), we can obtain other graphs with λ(G) ≥ 5 by applying a Y∆ operation to V9 or

10 00 V9 . Any of them contains a K6-minor, except if we apply Y∆ to vertex v1 (or equivalently, 0 to v5) of V9. So it could be true that λ(G) ≤ 4 if and only if G is obtainable from linklessly embeddable graphs by taking clique sums and subgraphs. Another open question is if λ(G) ≤ µ(G) for each graph G. That is, if for any representation L of G there is a matrix M satisfying (4) with dim(L) ≤ corank(M). This is true if µ(G) ≤ 4. In fact, a tempting, more general speculation is that for any natural number t:

(10) (???) a graph G satisﬁes λ(G) ≤ t if and only if G is obtainable from graphs H satisfying µ(H) ≤ t by taking clique sums and subgraphs (???) This has been proved for t ≤ 3, and the ‘if’ part for t ≤ 4.

References

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